@@ End of Day Wrap Tues 17/3/15 Sydney, Australia Time 11:15 pm ♪ ♫
Hi all,
Today our fantastic answerers were Alan, CPhill, MathsGod1, Shaomada, Themathematician7401, TayJay, Heureka, ThisGuy, Syrena and Bertie. Thanks all. Your blood's worth bottling. ლ(o◡oლ)
Interest Posts:
1) The Monty Hall problem. Yes again LOL I found a cool clip for it.
2) Finding volume of an odd shape. Melody
3) What is the last digit of pi? Thanks TayJay, anon and Heureka ![]()
4) Manipulating algebra Thanks CPhill and anon
5) Physics - Force and Power. Thanks Alan
6) Probability (secret santa) Melody
7) Fining an angle using sine rule Thanks melody and CPhill.
♫♪ ♪ ♫ ♬ ♬ MELODY ♬ ♬ ♫♪ ♪ ♫
Wed 18/3/15
1) A Serious weight problem. Thanks anon ![]()
2) Forum humour for the day. Thanks anon (oh and Chris lol)
3) Volume and SA, simultaneous equations. Thanks CPhill.
4) Find area when given the perimeter of a regular pentagon. Thanks Melody and CPhill
5) Diverge or converge that is the question. Thanks Melody CPhill and Alan
6) Complex numbers Thanks Heureka
7) Something for the younger students to ponder on. Thanks CPhill
8) What does mod really mean (again for the younger ones) Thanks Geno
9) solve sin x = cos x Thanks CPhill, Heureka and Melody
10) Simultaneous equations Thanks CPhill
11) Powers of complex numbers Thanks Heureka and anon
12) Synesthesia - I got off track but you might like the clips. LOL
♫♪ ♪ ♫ ♬ ♬ MELODY ♬ ♬ ♫♪ ♪ ♫
In a party,each of participants,Ada,Billy and 8 of their friends brings along a present for exchange.Each participant will draw a present randomly without displacement.
1)Given that Ada does ont get her own present,find the probability that Billy will get his own present.
I'll give it a shot. LOL
There are 10 people
the first person has 1 of 10 possible gifts.
the second person has 1 of 9 possible gifts
In total there are 10! possible combinations.
BUT
I might let Ada go first then Ada will have only 9 not ten gifts to choose from so the number of possibilities will be 9*9!
Now how many ways can billy get his own present.
I'm going to let billy go first and Ada go second
there is only one way that Billy can get his own gift
then there are 8 ways that Ada can get somone elses
etc
so maybe that will be 1*8*8!
So maybe
P(Billy gets his own given that Ada does not get her own) = $$\frac{8*8!}{9*9!}=\frac{8*1}{9*9}=\frac{8}{91}$$
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I hope that this is not total nonesence because that could be just a tad embarassing. LOL.
Don't worry - I can take it. :)) ¯\_(ツ)_/¯